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GR9677 #47
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Problem
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This problem is still being typed. |
Electromagnetism }Faraday Law
Recall Faraday's Law,  , where  . In words, this means that a changing flux (either a varying field or radius) induces a voltage.
The field is given as just B. The area of the loop is just  , i.e., the cross-sectional area of the cylinder. As the cylinder is spun around, its flux changes at the rate of N rps. The change in flux is thus  , and this is the magnitude of the potential difference in choice (C).
(Also, one can drop out the other choices from units. And, since the cylinder is moving in a magnetic field, the non-zero flux demands a voltage, so (A) can't be it.)
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Alternate Solutions |
ramparts
2009-09-26 12:01:14 | Units and limits are your friend. As yosun says, you can drop B and E because of units, and A for physical reasons, but that leaves D in addition to C. *Now* you take limits ;) As the number of coils goes to 0, does the voltage vanish or blow up? Not blowing up seems like a good bet - so pick C. |  |
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Comments |
South92
2009-10-20 07:52:11 | This solution is reasonable, but it seems strange that the charges on the wire do not feel the Lorentz force in the direction of the wire. Instead, the Lorentz force appears to point radially. |  | ramparts
2009-09-26 12:01:14 | Units and limits are your friend. As yosun says, you can drop B and E because of units, and A for physical reasons, but that leaves D in addition to C. *Now* you take limits ;) As the number of coils goes to 0, does the voltage vanish or blow up? Not blowing up seems like a good bet - so pick C. | | plapas
2009-04-01 13:17:02 | This excercise is tricky enough. The magnetic field does not change, thus the E =-dΦ/dt form cannot be used, since the magnetic flux through the section that is wounded around the cylinder does not change. However, the is an induced emf due to the rotation of the segment of the wire that is parallel to the one of the cylinder's base (that then continues along the axis of the cylinder - this section can be treated as a small rod rotating perpendicular to the lines of magnetic flux). Thus we can use the faraday law is the form E =Line Integral[(uxB) *dl], for the forementioned segment. If one thinks carefully, they can see that all other segments does not contribute because they lead to zero line integral... . The answer is (C).
Prologue
2009-10-20 23:21:57 |
Excellent catch! This is so true.
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Prologue
2009-10-20 23:54:00 |
Although I am not sure this is a form of Faraday's law. It looks to me like this:
Start with
Then do a path integral on both sides
Divide the equations by q
Then recognize that the left hand side is just an electric potential so that
Now we want to integrate just along that little kink that is flat against the end of the cylinder, the integral will go from 0 to R. The velocity of rotation and B are perpendicular and the lorentz force is along the wire so the force is along the wire, then dotted with dl (again along the wire so it is actually dr) it just gives
The negative sign is because of the opposition of our chosen dl to the lorentz force, in other words if we choose to go from 0 to R then our dl is in the opposite direction of the force on a positive ion in the wire.
But we can look at  as  and  is given by  where in the case  is  . So  where r is a variable. Then
Doing the integral you get
which is the answer apart from the negative sign, but you can just choose to do the path integral the opposite direction and it will be positive. In other words switch your voltmeter cords!
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carle257
2010-04-04 22:58:41 |
The standard  does work because flux is B x total area which is  (\pi R^2)) . Then the time derivative must also differentiate number loops as a function of time. This is given as N in the problem so the total flux derivative is  .
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| | sharpstones
2007-08-11 15:34:25 | Ah!! Now I understand!
The key here is that the wire is BEING wound around the rotating wooden cylinder during the whole time. So as it spins around more and more of the wire is wound around the cylinder and hence more flux is enclosed by the whole wire!
plapas
2009-08-05 10:04:20 |
THIS IS A WRONG APPROACH. THE EXERCISE DOES NOT MENTION THAT THE WIRE GRADUALLY WOUNDS AROUND THE CYLINDER. THERE IS SOME PORTION WOUNDED AROUND THE CYLINDER AND THE REST OF THE WIRE DIRECTS ALONG THE AXIS OF SYMMETRY. THE SOLUTION GIVEN ABOVE IS ALSO WRONG!!!
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Wildroses
2009-09-20 13:31:20 |
Actually, the problem does state that it is gradually being wound.rnrn"A wire is BEING wound around a rotating wooden cylinder".
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wittensdog
2009-10-30 19:43:44 |
all caps =/= correctness
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| | sharpstones
2007-04-05 05:33:49 | Yea, I don't understand either. As far as I can tell the flux is constant regardless of whether the cylinder is rotating or not. | | RebeccaJK42
2007-03-18 18:39:25 | Around which axis is the cylinder rotating? The way I pictured the problem, I don't see how the flux changes at all... | |
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